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Is there a field $F$ such that $F \cong F(X,Y)$ as fields, but $F \not \cong F(X)$ as fields?

I know only an example of a field $F$ such that $F$ isomorphic to $F(x,y)$ : this is something like $F=k(x_0,x_1,\dots)$. But in this case we have $F \cong F(x)$.

This is related to this question. This was previously asked on MSE but after 2 months and a bounty, no answer was provided.

Thank you very much!

Martin Sleziak
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Watson
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  • Related on MSE: https://math.stackexchange.com/questions/388291 – Watson Jan 03 '17 at 19:26
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    I have an example where I know $F\cong F(X,Y)$ but I don't know whether $F\cong F(X)$. There are torsion-free abelian groups $G$ such that $G\cong G\oplus\mathbb{Z}\oplus\mathbb{Z}$ but $G\not\cong G\oplus\mathbb{Z}$ (see http://mathoverflow.net/questions/218113/a-is-isomorphic-to-a-oplus-mathbbz2-but-not-to-a-oplus-mathbbz/227443#227443 for example). Let $\mathbb{Q}(G)$ be the field of fractions of the group algebra $\mathbb{Q}[G]$. Then $\mathbb{Q}(G)\cong\mathbb{Q}(G)(X,Y)$, but I don't see why it should be isomorphic to $\mathbb{Q}(G)(X)$. – Jeremy Rickard Jan 08 '17 at 21:34
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    Oops, I forgot to mention that I've posted a related question here: http://mathoverflow.net/questions/259117/the-field-of-fractions-of-the-rational-group-algebra-of-a-torsion-free-abelian-g – Jeremy Rickard Jan 08 '17 at 21:59

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